# NavList:

## A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Wayfinding

**Re: Silicon Sea Leg 72**

**From:**Dan Hogan

**Date:**2001 May 29, 18:35 EDT

On 29 May 2001, at 15:13, Noyce, Bill wrote: > > > > 4) What is the Compass Course(CC)/Course-to-Steer for Aruba Gap > > > > from the DR position? > > > > -- ------------------------------------------------------------ > > > > > > Using Law of Sines, I find I need to adjust my course North > > > about 1.5 degrees. > > > CC = 283.4 + 1.5 + 9 = 293.9d > > > > Mmmm..Barely OK. Try a current vector diagram. > > I drew a diagram to get the orientation of the pieces right, but I > have a hard time measuring the resulting course adjustment, because > it's so small. Thus the attempt at a 'digital' method. You can make the vector diagram any convenient size. > > TC=283.4, Set=265.0, angle between = 18.4 degrees > > sin 18.4 / speed = sin adj / drift > > sin 18.4 / 8.5 = sin adj / 0.7 > > I did this with log tables before; doing it on the computer now > says adj = 1.490 degrees. The diagram indicates it needs to be > added to the course. Is there something inherantly wrong with > this approach, or have I made a slip somewhere? Not being mathematically oriented I can't say, but in Shufeldts's The Calculator Afloat, he gives the following formulas: Finding correction angle and speed of advance Value of correction angle sin B = a x sin A /c Calculate speed of advance c = a x sin [d -(A + B)] /sin A Current Sailing when when track and speed of advance are specified Value of correction angle tan B = b x sin A / a - b x cos A Value of speed to steam a = b x sin A /sin B > > > > > > Mid-latitude using 13d 42' > > > dLat = 15' N dLon = 183d W dep = 177.8 W > > > TC = 274.6d dist = 178.3 nm > > > > OK. Be aware that above 500 miles at 90d/270d Mid-Latitude can give an > error. > > Above 1200 mils it gets unreliable. > > I saw a discussion initiated by Sam Chan about Mercator sailing > near 90 or 270, where the problem arises from rounding the course > before using dLat/cos(C) to compute the total distance. Is this > a similar problem? In Mid-Latitude sailing, where we already > have dLat and departure, the total distance can be computed either > as dLat/cos(C) or as departure/sin(C), so we don't have to divide > by a tiny rounded number. > > Or are you talking about the error in distance that arises from > assuming a spherical earth? What method would be > better? I assumed we weren't planning to sail a great-circle > course here. As I understand it it's caused by the formulas and the how they are derived. Mid-Latitude gets error prone when you cross the sperical quadrants. Try calculating using Mid-Latitude from North America to Australia and one N.AS. to South America of about equal distance. GC course comes after the Panama Canal transit ;^) Dan Hogan WA6PBY dhhogan{at}XXX.XXX Nav-L Web Page: http://www.wa6pby.com Catalina 27 "GACHA"